Question

A 1.5 kg 8-Ball is traveling at 2.3 m/s to the right when it strikes and bounces off a 5.2 kg bowling ball which is at rest. After the collision, the 8-Ball moves to the left with a velocity of 1.1 m/s. What is the final velocity of the bowling ball?
_______________m/s (nearest hundredth)

Answer 1

Answer:

-0.34m/s

Explanation

According to the law of conservation of momentum

m1u1 +m2 u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v s the common velocity

Given

m1 = 1.5kg

m2 = 5.2kg

u1 = 2.3m/s

u2 = -1.1m/s (left direction)

Required

Final velocity v of the ball

Substitute into the formula

1.5(2.3)-5.2(1.1) = (1.5+5.2)v

3.46-5.72 = 6.7v

-2.26 = 6.7v

v = -2.26/6.7

v = -0.34m/s

Hence the final velocity of the bowling ball is 0.34m/s towards the left